This course introduces students to the basic components of electronics: diodes, transistors, and op amps. It covers the basic operation and some common applications.

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Do curso por Georgia Institute of Technology

Introdução à Eletrônica

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This course introduces students to the basic components of electronics: diodes, transistors, and op amps. It covers the basic operation and some common applications.

Na lição

Diodes Part 2

Learning Objectives: 1. Examine additional applications of the diode. 2. Make use of voltage transfer characteristics to analyze diode circuit behavior.

- Dr. Bonnie H. FerriProfessor

Electrical and Computer Engineering - Dr. Robert Allen Robinson, Jr.Academic Professional

School of Electrical and Computer Engineering

Welcome back to Electronics, this is Dr. Robinson.

In this lesson, we're going to look at what are called Full-Wave Rectifiers.

In the previous lesson, you were introduced to diode half-wave rectifiers.

We didn't analyze the behavior of a particular half-wave rectifier circuit,

considering the diodes to be both ideal and non-ideal.

Our objectives for today are to introduce full-wave rectifiers.

We'll then examine their behaviors for sinusoidal inputs, and then,

analyze a diode full-wave rectifier circuit.

Now remember, our definition for

a rectifier is a non-linear device that modifies an input voltage,

such that the output voltage is greater than or less than, some threshold voltage.

Let's look at what that definition means graphically.

Here I've drawn a box to represent our full-wave rectifier.

We are applying as an input a sinusoidal varying voltage.

And I've drawn two possible outputs of the full-wave rectifier.

Let's look at this output first, and

determine it's relationship, with the input voltage.

You can see that when the input voltage is positive,

the output voltage is exactly equal to the input voltage.

But when the input voltage is negative,

the output voltage is equal to the input voltage multiplied by minus one.

This wave form is known as a positive full-wave rectified sine wave.

Positive because it's entirely greater than out threshold voltage of zero volts.

And full-wave, because the output is nonzero, for

every half period of the input.

Now remember for the half wave rectifier, our output was nonzero only for

every other half period of the input.

The output voltage waveform down here is known as a negative full-wave

rectified sine wave.

Negative because it's entirely below the threshold voltage of zero volts.

And full-wave rectified again, because the output is nonzero for

every half period of the input.

So here I've defined the behavior of a positive full-wave rectifier in terms of

rules that we can apply to the input.

If the input is greater than or equal to 0, the output is equal to the input.

However, if the inputs is less than 0,

the output is equal to the input times minus 1.

Now let's look at a circuit that may be used to implement a full-wave rectifier.

You may recognize this depology.

This is a wheatstone bridge depology,

where the elements of the bridge are diodes.

And in this case, the diodes are considered to be ideal.

A V in is a positive voltage, it will attempt to push current,

from its positive terminal, through the circuit, back to its negative terminal.

However, in this circuit, the paths the current can flow in,

are limited by the direction of the diodes.

So let's say current is,

is being pushed out of the positive terminal of the voltage source.

We can see that the current can't flow in this direction through D4 because of

its direction.

So the current must flow, through this branch, down through D1.

It's then forced to flow through the resistor because of the direction of D3.

Now at this node, we, if we just look at directions, it would appear that

current could flow both in this direction through D4, and this direction through D2.

However, we know that the voltage has decreased from this node,

by 1IR drop to get to this node.

Which means that this voltage is less than this voltage.

So D4 would be off, or reverse biased.

So the current that's coming through this branch through the resistor R,

must flow down through D2 and back to the minus side, of the input source.

So, when V in is positive, we can replace D1 and

D2 by short circuits, and D4 and D3 by open circuits.

And I've redrawn the circuit here, with those changes.

Now we can perform a similar analysis when V in is negative, and

arrive at this result.

Now the key to the operation of this circuit is to notice that, in this circuit

we know that current must be flowing in this direction, because of the diodes.

So current is flowing from the positive side of V out to the negative side of V

out which means V out has a positive voltage.

Now in this circuit, when V in is negative, remember for a negative

voltage we can consider this side to be positive, and this side negative.

Current must flow in this direction, because of the direction of the diodes.

And V out is again positive, because current is flowing from its

positive terminal to its negative terminal.

The two output voltage equations from our previous analysis can be

combined into a single piece-wise linear equation,

that governs the behavior of our original circuit.

We can see that V out is equal to V in, when V in is greater than or

equal to 0, and V out is equal to minus V in, when V in is less than 0.

This is the state where D1 and D2 are on, and D3 and D4 are off.

This is the state where D1 and D2 are off, but D3 and D4 are on.

Now another way of writing this equation is,

V out is equal to the absolute value of V in, for any conditions.

And for this reason a full-wave rectifier,

is also known as an absolute value circuit.

Now let's ask the question how would the behavior of

the full-wave rectifier change, if we consider the diode to be non-ideal, or

we included its forward voltage drop, modeling it like this.

So here is the full-wave rectifier drawn, assuming the diodes to be non-ideal.

We're also going to assume that the input voltage is positive,

such that current flows, from the positive side of V in, around the loop,

through D1, through the resistor R.

Through D2 and back to the mono side of the battery.

So under those conditions we know that D1 and D2 are both on, and D3 and

D4 are both off.

So here I've redrawn the circuit, replacing D3 and D4 by open circuits, and

replacing D1 and

D2 by batteries of values Vf, to represent their forward voltage drop.

Now we know that in this circuit, the current must be flowing around the loop.

In this direction.

Now remember, that when a diode is on, the current through the diode must flow from

the positive side of Vf to the negative side of Vf,

because from the anode to the cathode of the diode, we must have a voltage drop.

Now here I've redrawn the circuit, to make it somewhat simpler, or more compact.

Here are the two Vf sources representing the for,

forward voltage drop across diodes D1 and D2.

Now we know for

this circuit to apply, the current must be flowing around the loop in this direction.

From the positive side of the Vf, to the negative side of Vf.

However, the polarity of the batteries Vf,

is attempting to push current around the loop in this direction.

So, in order for there to be a net flow of current in this direction, for

this circuit to apply, we know that the input voltage must be greater than 2Vf.

But we can write an equation to see that.

Let's write that,

V in minus Vf plus Vf, divided by R,

is equal to the current, in this direction.

We can see that, V in, minus 2Vf,

out over R, is equal to the current.

And for this current to be positive,

we can see that V in must be greater than 2 Vf.

So I positive, implies that V in,

is greater than 2Vf.

So the condition under which this circuit applies, I've drawn down here,

as V in is greater than 2Vf.

Now, we can also use this circuit to determine,

the output voltage in terms of the input voltage.

We can see that V in, I'm going to

write a loop equation around this loop, we go up by V in, we go down by V,out.

We go down by Vf, we go down by Vf.

Which implies that, V out for this circuit,

is equal to V in minus 2Vf.

Now, we can perform a similar analysis, assuming that D3 and D4 are on, and D1 and

D2 are off.

The resulting circuit, is drawn here.

Now, for this circuit to apply,

we know that current must flow around the loop in this direction.

From the positive side of Vf to the negative side of Vf.

However, the polarity of the batteries is such that,

they are attempting to push current around the loop in this direction.

So for there to be a net flow of current in this direction,

we can see that V in must be negative, with its positive side here and

its negative side here, and that its magnitude must be greater than, 2Vf.

So, I've written here the condition, under which this circuit applies.

Now we could, like we did for the last circuit, write an equation to

determine this, but let's just use this condition by inspection.

Now, we can use this circuit to determine the output versus input

voltage relationship like before, again by writing a loop equation.

So we go up by V in, we go up by V out, we go up by Vf,

we go up by Vf, as we go around the loop.

From this, we can see that V out, is equal to negative V in-

Minus 2Vf.

Now we can take the two results from these previous analyses and

put them into a single piece-wise linear equation, that describes the behavior,

of the full-wave rectifier.

So here we can see that the output is equal to V in minus 2Vf when V

in is greater than 2Vf.

And that V out is equal to minus V in minus 2Vf when V

in is less than minus 2Vf.

Now if the input voltage, is a voltage between these two voltages of 2Vf and

minus 2Vf volts, then we know that, none of the diodes are on.

The current through the resistor would be 0 and

the output voltage would be equal to 0.

So, our full-wave rectifier considering the diodes to be

non-ideal really has three states.

The case where D1 and D2 are on, the case where D3 and D4 are on, and

the case where none of the diodes are on.

Now I've plotted this equation here on this

graph assuming a sinusoidal input voltage.

The input voltage is in blue, the output voltage is in red.

And also on this graph I've labelled,

the threshold voltages of 2Vf, and minus 2Vf.

Now you can see that, when the input voltage is greater than 2Vf volts,

in this region here, that we obtain the output voltage by subtracting, 2Vf.

So we subtract a voltage of 2Vf,

from the input voltage to obtain the output voltage.

Then, when then input voltage is less than minus 2Vf, we obtain the output

voltage by multiplying this voltage by minus 1, and then subtracting 2Vf.

So in other words, we flip that about the x axis, and then subtract 2Vf.

And when we do that we obtain a shape exactly the same as we obtain,

when the input voltage is greater than 2Vf.

Now in the region where the input voltage is between the two threshold voltages of

2Vf and minus 2Vf, we can see that the output voltage is 0.

So, let me leave you with some questions to think about.

If, in our full-wave rectifier circuit, the direction of

all four diodes was reversed, how would the behavior of the circuit change?

And, if you reversed the direction of any one of the diodes in the circuit,

how would the behavior of the circuit change?

So see if you can use our analysis, or

the methods we used during this lesson, to answer these questions.

So in summary, during this lesson,

we looked at the behavior of full-wave rectifiers.

And in our next lesson,

we're going to look what are called Voltage Transfer Characteristics, or

graphical depictions of the behavior of a nonlinear circuit.

Thank you and until next time.

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